यदि `A` `(-hati + 3hatj + 2hatk) , B (-4hati + 2hatj - 2hatk)` तथा `C(5hati + (lambda +1)hatj + mu hatk)` संरेखीय है तब :

Options

A. `lambda = 5, mu = 10`
B. `lambda = 10 , mu = 5`
C. `lambda = 4 , mu = 10`
D. `lambda = 5, mu = - 10`

Answer

To determine the values of \(\lambda\) and \(\mu\) such that the vectors \(A\), \(B\), and \(C\) are collinear, we need to set up the problem using the condition for collinearity of vectors. The vectors are given as:

\[ A = -\hat{i} + 3\hat{j} + 2\hat{k} \]
\[ B = -4\hat{i} + 2\hat{j} - 2\hat{k} \]
\[ C = 5\hat{i} + (\lambda + 1)\hat{j} + \mu\hat{k} \]

### Step 1: Form the determinant

For three vectors to be collinear, the determinant of the matrix formed by their components must be equal to zero. We can write the determinant as follows:

\[ \begin{vmatrix} -1 & 3 & 2 \\ -4 & 2 & -2 \\ 5 & \lambda + 1 & \mu \end{vmatrix} = 0 \]

### Step 2: Calculate the determinant

We will expand this determinant:

\[ = -1 \begin{vmatrix} 2 & -2 \\ \lambda + 1 & \mu \end{vmatrix} - 3 \begin{vmatrix} -4 & -2 \\ 5 & \mu \end{vmatrix} + 2 \begin{vmatrix} -4 & 2 \\ 5 & \lambda + 1 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \(\begin{vmatrix} 2 & -2 \\ \lambda + 1 & \mu \end{vmatrix} = 2\mu - (-2)(\lambda + 1) = 2\mu + 2\lambda + 2\)

2. \(\begin{vmatrix} -4 & -2 \\ 5 & \mu \end{vmatrix} = (-4)(\mu) - (-2)(5) = -4\mu + 10\)

3. \(\begin{vmatrix} -4 & 2 \\ 5 & \lambda + 1 \end{vmatrix} = (-4)(\lambda + 1) - (2)(5) = -4\lambda - 4 - 10 = -4\lambda - 14\)

Now substituting back into the determinant expansion:

\[ = -1(2\mu + 2\lambda + 2) - 3(-4\mu + 10) + 2(-4\lambda - 14) = 0 \]

### Step 3: Simplify the equation

Expanding this gives:

\[ -2\mu - 2\lambda - 2 + 12\mu - 30 - 8\lambda - 28 = 0 \]

Combining like terms:

\[ (12\mu - 2\mu) + (-2\lambda - 8\lambda) + (-2 - 30 - 28) = 0 \]

This simplifies to:

\[ 10\mu - 10\lambda - 60 = 0 \]

### Step 4: Rearranging the equation

Rearranging gives:

\[ 10\mu - 10\lambda = 60 \]

Dividing through by 10:

\[ \mu - \lambda = 6 \]

### Step 5: Conclusion

Thus, we have the relationship between \(\mu\) and \(\lambda\):

\[ \mu = \lambda + 6 \]